Strengths

Footing

Failures

q

1.Bearing failure q A  ult

FS

2.Serviceability failure

q

3.Excessive total settlement

P





M

Centroidal Axis









P/A



My/I

Kern









B









L

L/6 L/6 eK 

6

L P My

q 

A I

Pc=0







240 lb/ft2





300 lb/ft2





540 lb/ft2

Pc







2 ft of soil @120 lb/ft3 240 lb/ft2





300 lb/ft2

2 ft of concrete @150 lb/ft3



540 lb/ft2

Gross soil pressure =

540+qn lb/ft3 qn=Pc/A

P



M









Net soil pressure

qn=Pc/A

Only Dead load and Live load



DStructure, footing , surch arg e   L

Area of footing A

qa



Including Wind load also

DStructure, footing , surch arg e   L  W

Area of footing A

1.33 q a





1.4 Dstructure   1.7 L

qnu 

A

or



0.75[1.4 Dstructure  1.7 L  1.7W ]

qnu 

A

Isolated footing with axial load and moment





Design Input



•Grade of concrete

•Grade of steel

•Allowable soil pressure

•Column load

•Moment

•Condition about overburden of soil

Design procedure

Checking the footing area

•Assume column size

•Assume footing size

•Total load=wt of footing+column load

•Average soil pressure q= total load / footing area

Allowable soil pressure > Average soil pressure

then footing are is ok

Finding factored loads



•Factored column load=1.4xdead load+1.7xlive load



•Average factored soil pressure qnet=Factored column load/footing area





•Average dv=total thick – cover - bar dia

•Eccentricity e=Mu/Pu



Solving two equations to get q1 and q2

•Service load equation

• Vertical equilibrium

Check for One way shear

•Take the critical section as ‘d’ distance from the

face of the column (as per ACI 318 )

1.find pressure at critical section

2.find Vu

3.find  Vu

If Vu>Vu

Check for Two way shear

• Take the critical section as ‘d/2’ distance all around

the face of the column (as per ACI 318 )

1. find pressure at critical section

2. Find the value of bo

2.find Vu

3.find Vu is smallest of these

 4 '

Vc    2   f c bo d



 c 



 d 

Vc    s  2  f c' bo d

 b 

 o 



Vc   4 f c' bo d



Vu  Vc Safe against two way shear



Otherwise increase the depth

Check for flexural reinforcement

Max moment occurs at the face of the column





•Find average pressure

•Find required moment Mn=Mu/ 

•Find required Ast

•Min Ast=0.0018bh

•Max spacing=18 in

Development length

0.04 Af y

ld 

f c'

Thank You